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Chap. 11 Graph Theory and Applications. Directed Graph. (Undirected) Graph. Vertex and Edge Sets. Walk. Closed (Open) Walk. Trail, Path, Circuit, and Cycle. Comparison of Walk, Trail, Path, Circuit, and Cycle. Theorem 11.1. Observation:. Theorem 11.1. It suffices to show from a to b,
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Theorem 11.1 Observation:
Theorem 11.1 • It suffices to show from a to b, • the shortest trail is the shortest path. • 2. Let be the shortesttrail from a to b. • 3. • 4.
Connected Graph connected graph disconnected graph
Induced Subgraph Which of the following is an induced subgraph of G? O O X
Components of a Graph connected sugraph 1 2
Isomorphic Graphs Which of the following function define a graph isomorphism for the graphs shown below? X O
Isomorphic Graphs Are the following two graphs isomorphic? X In (a), a and d each adjacent to two other vertices. In (b), u, x, and z each adjacent to two other vertices.
a b d c
Theorem 11.3 (⇒) 1. 2. 3. 4. 5. 6. 7.
Theorem 11.3 8. 9.
Theorem 11.3 (⇐) 1. 2. 3.
Theorem 11.3 4. 5. 6. 7. 8.
Theorem 11.3 9. 10. 11. 12. 13. 14.
Corollary 11.2 (⇐) 1. 2. 3. 4. (⇒) The proof of only if part is similar to that of Theorem 11.3 and omitted.
Theorem 11.4 The proof is similar to that of Theorem 11.3 and omitted.
Planar Graph Which of the following is a planar graph? O O
Euler’s Theorem v = e = r = v – e + r = 2 7 8 3
Euler’s Theorem • Proof. 1. Use induction on v (number of vertices). • 2. Basis (v = 1): • G is a “bouquet” of loops, each a closed curve in the embedding. • If e = 0, then r = 1, and the formula holds. • Each added loop passes through a region and cuts it into 2 regions. This augments the edge count and the region count each by 1. Thus the formula holds when v = 1 for any number of edges.
e Euler’s Theorem • 3. Induction step (v>1): • There exists an edge e that is not a loop • because G is connected. • Obtain a graph G’ with v’ vertices, e’ edges, and r’ regions by contracting e. • Clearly, v’=v–1, e’=e–1, and r’=r. • v’– e’+ r’ = 2. • Therefore, v-e+r=2. (induction hypothesis)
Corollary 11.3 If G is a simple planar graph with at least three vertices, then e≤3v–6. (A simple graph is not a multigraph and does not contain any loop.) If also G is triangle-free, then e ≤ 2v–4. 1. It suffices to consider connected graphs; otherwise, we could add edges. 2. If v 3, every region contains at least three edges (L(Ri) 3r). 3. 2e=L(Ri), implying 2e3r. 4. By Euler’s Theorem, v–e+r=2, implying e≤ 3v– 6. (L(Ri) 4r) (2e4r) (e≤ 2v–4)
K5 (e = 10, n = 5) K3,3 (e = 9, n = 6) Nonplanarity of K5 and K3,3 • These graphs have too many edges to be planar. • For K5, we have e = 10>9 = 3n-6. • Since K3,3 is triangle-free, we have e = 9>8 = 2n-4.
